Given an array nums and a value val, remove all instances of that value in-place and return the new length.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
The order of elements can be changed. It doesn't matter what you leave beyond the new length.
Example 1:
Given nums = [3,2,2,3], val = 3, Your function should return length = 2, with the first two elements of nums being 2. It doesn't matter what you leave beyond the returned length.
Example 2:
Given nums = [0,1,2,2,3,0,4,2], val = 2,
Your function should return length = 5
, with the first five elements of nums containing 0,1,3,0,4
Note that the order of those five elements can be arbitrary.
It doesn't matter what values are set beyond the returned length.
给一个数组[1,2,4,1,2] 去除指定元素val=2 为[1,4,1]并返回长度3
public int removeElement(int[] nums, int val) {
int index=0;
for(int num:nums)
{
if(num!=val)
{
nums[index++]=num;
}
}
return index;
}
Implement strStr()
Implement strStr().
Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
Example 1:
Input: haystack = "hello", needle = "ll" Output: 2
Example 2:
Input: hays方法二
tack = "aaaaa", needle = "bba"
Output: -1
方法一
用自带方法
public int strStr(String haystack, String needle) {
return haystack.indexOf(needle);
}
方法2
public int strStr(String haystack, String needle) {
for(int i=0;;i++)//haystack字符索引
{
for(int j=0;;j++)//needle字符索引
{
if(j==needle.length())return i;//如果needel字符比较到头则返回haystack此时索引
if(i+j==haystack.length()) return -1;//如果超出范围则返回-1
if(needle.charAt(j)!=haystack.charAt(i+j)) break;//依次比较字符是否相同
}
}
}